I have a question about finding an approximate value $x$ for the following expression:
$$\frac{(e^{x\alpha_{1}})^2 + (e^{x\alpha_{2}})^2 + \ldots + (e^{x\alpha_{n}})^2}{\displaystyle \left(\sum_{i = 1}^{n} e^{x\alpha_{i}}\right)^2} = \beta,$$
where $\alpha_{i}$ and $\beta$ are constant values. There is a closed form solution for $n=2$, but I'm pretty sure there's no closed form solution for any $n$.
A graph of the equation is shown below ($\beta = 0.4$):
Is there any way I can approximate the value of $x$ in which the expression to the right will give me a value somewhat near $\beta$? I am not too familiar with methods of numerical approximations - how could I do this?
Let's say I have a computer to help me solve this. Is there a better way than simply trying all values of $x$ from $-20$ to $20$ with step $0.1$?
This function has a local minimum at $x=0$. In the positive $x$, it is asymptotic to
$$\frac{e^{2\alpha_n\infty}}{e^{2\alpha_n\infty}}=1$$ assuming that the $\alpha$ are in increasing order.
In the negatives,
$$\frac{e^{-2\alpha_1\infty}}{e^{-2\alpha_1\infty}}=1.$$
It is likely that the function is monotonic on both sides, though I couldn't prove it. Then if $\beta<\dfrac1n$, there is no solution. Otherwise, look for $x$ such that $f(x)>\beta$. You can do this by trying values from $\alpha_nx=1$ (resp. $-\alpha_1x=1$) and doubling $x$ until you establish $f(x)>\beta$.
Now you have two points on either sides of $y=\beta$ and you can start using one of the secant or regula falsi methods.